# Commutator of two subgroups is normal in join

Jump to: navigation, search

## Statement

Suppose $H, K$ are subgroups of a group $G$. Then, the Commutator of two subgroups (?) $[H,K]$ is a normal subgroup of the Join of subgroups (?) $\langle H, K \rangle$.

## Facts used

1. Subgroup normalizes its commutator with any subset: If $H \le G$ is a subgroup and $A$ is a subset of $G$, then $H$ normalizes the following subgroup:

$[A,H] = \langle [a,h] \mid a \in A, h \in H \rangle$

Here, $[a,h] = a^{-1}h^{-1}ah$ is the commutator of the two elements.

## Proof

Given: Two subgroups $H, K \le G$.

To prove: $[H, K ] \triangleleft \langle H, K \rangle$.

Proof: By fact (1), $H$ normalizes $[K, H]$, which is the same as $[H,K]$. Also, $K$ normalizes $[H,K]$. Thus, the normalizer of $[H,K]$ in $G$ contains both $H$ and $K$, hence it contains $\langle H, K \rangle$, proving that $[H,K]$ is normal in $\langle H, K \rangle$.